In the expansion of `(1+x)^(2)(1+y)^(3)(1+z)^(4)(1+w)^(5)`, the sum of the coefficient of the terms of degree 12 is :

Options

A. 61
B. 71
C. 81
D. 91

Answer

To solve the problem, we need to find the sum of the coefficients of the terms of degree 12 in the expansion of \( (1+x)^2(1+y)^3(1+z)^4(1+w)^5 \).

### Step-by-Step Solution:

1. **Identify the total degree**: The total degree of the expression is given by the sum of the degrees of each factor:
- \( (1+x)^2 \) contributes a maximum degree of 2,
- \( (1+y)^3 \) contributes a maximum degree of 3,
- \( (1+z)^4 \) contributes a maximum degree of 4,
- \( (1+w)^5 \) contributes a maximum degree of 5.

Therefore, the maximum degree is \( 2 + 3 + 4 + 5 = 14 \).

2. **Set up the problem**: We need to find the sum of the coefficients of the terms that have a total degree of 12. This means we need to find combinations of terms from each factor that add up to 12.

3. **Change variables**: To simplify the calculation, we can substitute \( x = y = z = w = x \). This gives us:
\[ (1+x)^2(1+x)^3(1+x)^4(1+x)^5 = (1+x)^{2+3+4+5} = (1+x)^{14} \]

4. **Find the coefficient of \( x^{12} \)**: We need to find the coefficient of \( x^{12} \) in the expansion of \( (1+x)^{14} \). The coefficient of \( x^k \) in \( (1+x)^n \) is given by \( \binom{n}{k} \).

Here, we need \( n = 14 \) and \( k = 12 \):
\[ \text{Coefficient of } x^{12} = \binom{14}{12} \]

5. **Calculate \( \binom{14}{12} \)**: Using the property of binomial coefficients, we know that \( \binom{n}{k} = \binom{n}{n-k} \). Thus:
\[ \binom{14}{12} = \binom{14}{2} \]

Now calculate \( \binom{14}{2} \):
\[ \binom{14}{2} = \frac{14 \times 13}{2 \times 1} = \frac{182}{2} = 91 \]

6. **Final answer**: The sum of the coefficients of the terms of degree 12 in the expansion is \( 91 \).